Một sự mở rộng mới của phương pháp thác triển theo tham số giải phương trình toán tử loại hai

No.21_June 2021 |p.157-165 157 TẠP CHÍ KHOA HỌC ĐẠI HỌC TÂN TRÀO ISSN: 2354 - 1431 A NEW EXTENSION OF PARAMETER CONTINUATION METHOD FOR SOLVING OPERATOR EQUATIONS OF THE SECOND KIND Ngo Thanh Binh1,* 1 Nam Dinh University of Technology Education, Vietnam *Email address: ntbinhspktnd@gmail.com https://doi.org/ 10.51453/2354-1431/2021/521 Article info Abstract: Recieved: 27/3/2021 Accepted: 3/5/2021 In this paper, we propose an extension of the parameter continuation method for solving operator equations of the second kind. By splitting of the operator into a sum of two operators: one monotone, Lipschitz-continuous and one contractive, the applicability of the method is broader. The suitability of the proposed approach is presented through an example. Keywords: Parameter continuation method, Monotone operator, Contractive operator, Operator equations of the second kind, Approximate solution. No.21_June 2021 |p.157-165 158 TẠP CHÍ KHOA HỌC ĐẠI HỌC TÂN TRÀO ISSN: 2354 - 1431 MỘT SỰ MỞ RỘNG MỚI CỦA PHƯƠNG PHÁP THÁC TRIỂN THEO THAM SỐ GIẢI PHƯƠNG TRÌNH TOÁN TỬ LOẠI HAI Ngô Thanh Bình1,* 1 Trường Đại học Sư phạm Kỹ thuật Nam Định, Việt Nam *Địa chỉ email: ntbinhspktnd@gmail.com https://doi.org/ 10.51453/2354-1431/2021/521 Thông tin bài viết Tóm tắt Ngày nhận bài: 27/3/2021 Ngày duyệt đăng: 3/5/2021 Trong bài báo này, chúng tôi đề xuất một mở rộng mới của phương pháp thác triển theo tham số giải phương trình toán tử loại hai. Bằng cách tách toán tử thành tổng của hai toán tử: toán tử đơn điệu, liên tục Lipschitz và toán tử co, khả năng áp dụng của phương pháp được mở rộng. Sự phù hợp của cách tiếp cận đề xuất được trình bày thông qua một ví dụ. Từ khóa: Phương pháp thác triển theo tham số, Toán tử đơn điệu, Toán tử co, Phương trình toán tử loại hai, Giải xấp xỉ. 1. Introductions Parameter continuation method (PCM) was suggested and developed by Bernstein [1] and Schauder [3] which is the inclusion of the equation ( ) 0P x  into the one-parametric family of equations ( , ) 0, [0,1]G x    connecting the given equation ( 1)  with a solvable equation ( 0)  and study the dependence of the solution from parameter. The PCM is a powerful technique for solving operator equations, see for example [5– 7]. Gaponenko [2] introduced the PCM for solving operator equations of the second kind ( ) ,x A x f  (1) where A is a Lipschitz-continuous and monotone operator, which operates in an arbitrary Banach space .X The monotone operator in Banach space is defined as follows. Definition 1.1. [2, Definition 2] The mapping A , which operates in the Banach space X is called monotone if for any elements 1 2,x x X and any 0  the following inequality holds  1 2 1 2 1 2( ) ( ) .x x A x A x x x     (2) Remark 1.1. [2, Remark 1] If X is Hilbert space then the condition of monotonicity (2) is equivalent to the classical condition 1 2 1 2 1 2( ) ( ), 0, , ,A x A x x x x x X     N.T.Binh/ No.21_Jun 2021|p.157-165 159 where ,  is an inner product in the Hilbert space X . We obtain the following result from the definition above. Lemma 1.1. [2, Lemma] Assume that A is a monotone mapping which operates in the Banach space .X Then for any elements 1 2,x x X and any positive numbers 1 2 1 2, , 0 1      , the following inequality holds    1 2 1 1 2 1 2 2 1 2( ) ( ) ( ) ( ) .x x A x A x x x A x A x        The results obtained by Gaponenko are summarized in Theorem 1 and Theorem 2. Theorem 1.1. [2, Theorem 1] Suppose that the mapping ,A which operates in the Banach space X is Lipschitz - continuous and monotone. Then the equation (1) has a unique solution for any element f X . The following iteration process is constructed to find approximate solutions of the equation (1).  1 1 1 1 ( ) ( ) ( ) , , ,..., 0,1, ... . 3i i j p N terms x A x A x A x f i j p N N N         The symbolic notation (3) should be understood as the following iteration processes, which consist of N iteration processes         (1) 1 0 (1) (2)1 (1) 0 11 ( 1) 1 1 ( 1) 0 1 11 ( ) , 0, 1, 2 d ,..., ( ) , 0, 1, 2,... , 4a 4b 4 , . 4c. . .. 2 ., ( ) 0,1, ,. . i i j jj l N N N pp x A x x i x AG x x j x AG G x f p                        For simplicity, assume that (0) 0A  and the number of steps in each iteration scheme of the iteration process (3) is the same and equals n . Denoting ( , ) nx n N x as the approximate solutions of the equation (1), which is constructed by the iteration process (3). In this case, Gaponenko received the error estimations of approximate solutions of the equation (1), which are presented in the following theorem. Theorem 1.2. [2, Theorem 2] Assume that the conditions of Theorem 1.1 are satisfied. Then the sequence of approximate solutions { ( , )}, 1, 2, ...x n N n  constructed by iteration process (3) converges to the exact solution x of the equation (1). Moreover, the following estimates hold 1 1 ( , ) . 1 1 n qN q q e x n N x f q e       (5) where L is Lipschitz coefficient of the operator ,A N is the smallest natural number such that 1, 1, 2, .. .. L q n N    2. MAIN RESULTS Consider the general operator equation (1) ( ) ,x A x f  where A is a nonlinear operator from a Banach space X into ,X f is a given function in .X Without loss of generality, one can express the operator A as a composition of two operators 1A and 2.A Then the equation (1) can be rewritten as follow 1 2( ) ( ) .x A x A x f   (6) Theorem 2.1. Assume that 1A is a Lipschitz- continuous and monotone operator, 2A is a contractive operator. Then the equation (1) has a unique solution. Proof. We take a minimal natural number N such that 1 0 0 1 1,q L N     , where L is the Lipschitz coefficient of the operator 1A . The equation (6) can be written as the following form 0 1 2( ) ( ) .x N A x A x f   (7) Consider the following subsidiary problems. Problem 1 ( 1N  ). Consider the operator equation 0 1 2( ) ( ) .x A x A x f   (8) We shall carry out a change of variable (1) 0 1 1( ) ( ).x x A x G x   (9) We have N.T.Binh/ No.21_Jun 2021|p.157-165 160 0 0 0 1 1 1( ) ( ) , , . x x L x A q A x x x x x X          Hence 0 1A is a contractive operator with contraction coefficient equal to 1 0 1q L  . Then the equation (9) has a unique solution for any (1)x X , i.e., the operator 1 (1)1 ( )G x  is determined in the whole space .X By virtue of the monotonicity of the operator 1,A the operator 1 1G  is Lipschitz - continuous with Lipschitz coefficient equal to 1 . Indeed, for any (1) (1),x x X , we have   1 (1) 1 (1) 1 1 0 1 1 (1) (1) ( ) ( ) ( ) ( ) . G x G x x x x x A x A x x x            After changing the variable (9), the equation (8) will take the following form  (1) (1) 1 (1)1 12 ( ) .P x x A G x f   (10) For any (1) (1),x x X , we have 1 (1) 1 (1) 1 1 1 (1) 1 (1) (1) (1) 2 1 1 2 2 2 ( ) ( ) ( ) ( ) , A G x A G x q G x G x q x x         where 2q is contraction coefficient of the operator 2.A Thus 1 2 1A G  is a contractive operator with contraction coefficient equal to 2 1q  . Then the equation (10) has a unique solution for any .f X Consequently, the equation (8) has a unique solution 0( )x  for any .f X Problem 2 ( 2N  ). Consider the operator equation 0 1 22 ( ) ( ) .x A x A x f   (11) We shall carry out two changes of variables     (1) 0 1 1 (2) (1) 1 (1) (1) 0 1 1 2 ( ) ( ), 12a ( ) ( ). 12b x x A x G x x x A G x G x          For any (1) (1),x x X , we have 1 (1) 1 (1) (1) (1) 0 1 1 0 1 1 0 (1) (1) 1 ( ) ( ) . A G x A G x L x x q x x         Hence 1 0 1 1A G  is a contractive operator with contraction coefficient equal to 1 1q  . Then the equation (12b) has a unique solution for any (2)x X , i.e., the operator 12G  is determined in the whole space .X By Lemma 1.1, for any (2) (2),x x X , we have 1 (2) 1 (2) (1) (1) 2 2 0 1 1 0 1 1 ) ) [ ( ) ( )] 2 [ ( ( ( ) )] (G x G x x x x x A x A x x x A x A x              (1) (1) 1 (1) 1 (1) (2) (2) 0 1 1 1 1[ ) )] .( (x x A G x A G x x x        Thus the operator 1 2G  is Lipschitz - continuous with Lipschitz coefficient equal to 1 . After changing the variables (12a) and (12b), the equation (11) will take the following form (2) (2) 1 1 (2)2 2 1 2( ) ( ) .P x x A G G x f     (13) For any (2) (2),x x X , we have 1 1 (2) 1 1 (2) (2) (2) 2 1 2 2 1 2 2( ) ( ) . A G G x A G G x q x x       Thus 1 1 2 1 2A G G   is a contractive operator with contraction coefficient equal to 2 1q  . Then the equation (13) has a unique solution for any f X . Therefore the equation (11) has a unique solution 0(2 )x  for any f X . Problem N ( 2N  ). Consider the operator equation 0 1 2 1 2( ) ( ) ( ) ( ) .x N A x A x x A x A x f      (14) We shall carry out N changes of variables         (1) 0 1 1 (2) (1) 1 (1) (1) 0 1 1 2 ( ) ( 1) 1 1 ( 1) ( 1) 0 1 1 1 ( ) ( ), 15a ( ) ( ), 15b ..., 15c ( ) ( ). 15dN N N NN N x x A x G x x x A G x G x x x A G G x G x                    Similarily, we show that the operators 1 3 4 1 1,, ..., NG G G    are determined in the whole space X and are Lipschitz - continuous with Lipschitz coefficients equal to 1 . Hence after the change of variables (15a)-(15d) the equation (14) will take the following form N.T.Binh/ No.21_Jun 2021|p.157-165 161 ( ) ( ) 1 1 ( ) 2 1( ) ( ) . N N N N NP x x A G G x f     (16) For any ( ) ( ),N Nx x X , we have 1 1 ( ) 1 1 ( ) 2 1 2 1 ( ) ( ) 2 ( ) ( ) . N N N N N N A G G x A G G x q x x       Thus 1 1 2 1 NA G G   is a contractive operator with contraction coefficient equal to 2 1q  . Then the integral equation (16) has a unique solution for any f X . Consequently, the equation (14) has a unique solution 0( )x N x X   for any f X , i.e. the equation (1) has a unique solution x X for any f X .This completes the proof. We now construct the iterative algorithm to find approximate solution of the operator equation (1). Firstly, we construct the iterative algorithm to find approximate solution of the Problem 1 . The approximate solutions of the equation (10) are obtained by using the standard iteration process (1) (1)1 (1) 2 11 0( ) , 0,1,2,..., .jjx A G x f j x f        At the same time at each step of above iteration process when calculating the value 1 (1)1 ( )jG x  we will again use the standard iteration process (1) (1) 1 0 1 0( ) , 0,1,2,..., .i i j jx A x x i x x      As a result, the approximate solutions of the equation (8) can be found by the following iteration processes     (1) 1 0 (1) (1)1 ( 1 1 1) 1 02 ( ) , 0,1,2,..., 17a ) , 0,1,2,..., . 17b( i i j jj x x x i x A G x f j x f A           Next, we construct the iterative algorithm to find approximate solution of the Problem 2. The approximate solutions of the integral equation (13) are obtained by using the standard iteration process (2) (2) (2)1 1 2 1 2 01 ( ) , 0,1,2,..., .l lx A G G x f l x f         At the same time we will use “subsidiary” iteration processes to invert the operators 1 2,G G at each step of this iteration process when calculating the value of (2)1 1 1 2 ( ).lG G x   Hence the approximate solutions of the integral equation (11) can be found by iteration processes       (1) 1 0 1 (1) (2)1 (1) 0 1 11 (2) (2) (2)1 1 2 1 2 01 ( ) , 0,1,2,..., 18a ( ) , 0,1,2,..., 18b ( ) , 0,1,2,..., . 18c i i j jj l l l x A x x i x A G x x j x A G G x f l x f                      Finally, we construct the iterative algorithm to find approximate solution of the Problem .N The approximate solutions of the integral equation (16) are obtained by using the standard iteration process ( ) ( )1 1 ( ) 2 11 0( ) , 0,1,2,..., . N NN N ppx A G G x f p x f         At the same time we will use “subsidiary” iteration processes to invert the operators 1 2, ,..., NG G G at each step of this iteration process when calculating the value of 1 1 1 ( )1 2 ( ) N N pG G G x    . Hence the approximate solutions of the equation (14) can be found by iteration processes         (1) 1 0 1 (1) (2)1 (1) 0 1 11 ( ) ( )1 1 ( ) 2 11 0 ( ) , 0,1,2,..., 19a ( ) , 0,1,2,..., 19b ..., 19c ( ) , 0,1,2,..., . 19d i i j jj l N NN N pp x A x x i x A G x x j x A G G x f p x f                      The iteration processes (19a)-(19d) can be written as the following symbolic notation   1 1 1 1 2 1 1 1 ( ) ( ) ( ) ( ) , , ,..., 0,1,... . 20 i i j h p N terms x A x A x A x A x f N N N i j p          Assume that the number of steps in each iteration scheme of iteration processes (19a)-(19d) is the same and equals n . Let nx be approximate solutions of the equation (1). Note that nx depends on N . Hence we denote ( , ) nx n N x . We have the following theorem. Theorem 2.2. Let the assumptions of Theorem 2.1 be satisfied. Then the sequence of approximate solutions { ( , )}, 1,2,...x n N n  constructed by iteration processes (19a)-(19d) converges to the exact solution x X of the operator equation (1). N.T.Binh/ No.21_Jun 2021|p.157-165 162 Moreover, the following estimates hold 1 1 1 1 11 2 2 2 1 2 11 1 ( , ) , 1 1 1 1 n n q N n q q q e x n N x q f q q q e                 (21) where N is the smallest natural number such that 1 1 L q N   , L is Lipschitz coefficient of the operator 1 2,A q is a contraction coefficient of the operator 2 , 1,2,....A n  . Proof. Without loss of generality, we assume that 1 2(0) 0, (0) 0.A A  Let us consider successive problems 1, 2, ..., .N The approximate solutions of Problem 1 assumes are obtained by iteration processes (17a)-(17b). The values 1 (1)1 ( )jG x  are calculated by using the iteration process (17a) with the error 1 * (1)1 1 . 1 n n j q x x x q     For any  1,2,...,k n , we have (1) (1) (1) (1)1 1 2 1 2 11 1 2 1(1) (1) (1) (1) 2 2 1 01 2 ( ) ( ) , k k k k k k k x x A G x A G x q x x q x x                 so that   (1) (1) (1) (1)(1) (1) 1 1 0 0 1 2 (1) (1) (1) 22 2 1 0 0 (1) (1) (1)2 1 0 0 2 1 1 . 1 j j j j j j x x x x x x q q q x x x q x x x q                      Since 1(0) 0A  , we have 1 0 1(0) 0 (0) 0G A   . Hence (1) (1) (1) (1)1 2 11 0 0 0 1 1 2 1 2 1 2 ( ) ( ) (0) . x x A G x f x A G f A G q f           Then from above inequality it follows that (1) (1) (1)(1) 2 2 21 0 0 2 2 1 2 2 2 2 2 1 1 1 1 1 1 . 1 1 j j j n n q q x x x x q f f q q q q q f f f q q                  Consequently, the values 1 (1)1 ( )jG x  are calculated with the error * 1 1( ) ( ) ( ),nn n x x n      where 11 1 2 1 2 1 ( ) . 1 1 nn qq n f q q       (22) Since 2A is a contractive operator with contraction coefficient equal to 2 1q  , the error 1( )n in specifying the argument of the operator 2A is equivalent to the error 2 1( )q n in specifying the right - hand side f of the integral equation (8). On the other hand, the operator 1 1P  is Lipschitz - continuous with Lipschitz coefficient equal to 2 1 1 q . Indeed, for any ,f f X , we have 1 1 (1) (1) 1 1 (1) (1) 1 (1) 1 (1) 1 (1) 1 (1) 2 1 2 1 2 1 2 1 ( ) ( ) ( ) ( ) ( ) ( ) P f P f x x x x A G x A G x A G x A G x                   (1) (1) 1 (1) 1 (1) 1 (1) 1 (1) 2 1 2 1 2 1 2 1( ) ( ) ( ) ( )x x A G x A G x A G x A G x          (1) (1) (1) (1) 1 1 2 (1) (1) 2 ( ) ( ) , P x P x q x x f f q x x         so that 1 1 1 1 2 .( ( 1 ) ) 1 P f P f f f q      Hence the substitution of the error 2 1( )q n into the right – hand side of the integral equation (8) causes an error of not more than 2 2 1( ) 1 n q q   in the corresponding solution (1)x . The error of an iteration process in the calculation of (1)x equals 1 2 21 nq f q   . Therefore we have 1 (1) (1) 2 2 1 2 2 ( ) . 1 1 n n q q x x n f q q        The inverse substitution, i.e., the transition from the variable (1)x to the variable x again introduces the error 1( )n . Then the error of approximate solutions nx of Problem 1 gives the estimate N.T.Binh/ No.21_Jun 2021|p.157-165 163 1 2 2 0 1 1 2 2 1 2 1 2 2 ( ) ( ) ( ) 1 1 1 ( ) . 1 1 n n n q q x x n n f q q q n f q q                 The approximate solutions of Problem 2 assumes are obtained by iteration processes (18a)-(18c). The values (2)1 1 1 2 ( )lG G x   are calculated by using iteration processes (18a)-(18b). The values 1 (1) 1 ( )jG x  are calculated by using the iteration process (18a) with the error 1 * (1)1 1 . 1 n n j q x x x q     We have (1) (1) (1) (1)(1) (1) 1 1 0 0 .j j jx x x x x x      Since the operator 12G  is Lipschitz - continuous with Lipschitz coefficient equal to 1 , it follows that   (1) (1) (2) (2)1 1 2 21 1 (2) (2) 1 ) ) , 1,2,..., . ( (k k k k k k x x G x G x x x k n             Hence (2) (2) (2) (2)(1) (2) 1 1 0 0 .j j jx x x x x x      For any  1,2,...,k n , we have (2) (2) (2) (2)1 1 1 1 1 1 2 1 1 21 1 2 1(2) (2) (2) (2) 2 2 1 01 2 ( ) ( ) . k k k k k k k x x F G G x F G G x q x x q x x                   Thus   (2) (2) (2)(1) 1 22 2 2 1 0 0 (2) (2) (2)2 1 0 0 2 1 1 . 1 j j j j x q q q x x x q x x x q              Since 1(0) 0,A  it follows that 11 00) 0 0) 0( (G A   and 1 2 0 1 1(0) 0 0 0( )AG G    . Therefore (2) (2) 1 1 2 1 21 0 1 1 1 1 2 1 2 2 1 2 2 ( ) ( ) (0) . x x A G G f f f A G G f A G G q f              Then we have (2) (2) (2)(1) 2 1 0 0 2 2 2 2 2 2 2 1 2 2 1 1 1 1 1 1 1 . 1 j j j n n q x x x x q q q f f q q q f f q q f q                + + Therefore the values 1 (1)1 ( )jG x  are calculated with the error 1 1 * 1 2 1 2 1 ( ). 1 1 n n n q q x x f n q q         Since 0 1A is a contractive operator with contraction coefficient equal to 1 1q  , the error ( )n in specifying the argument of the operator 0 1A is equivalent to the error 1 ( )q n in specifying the right - hand side (2)x of the equation (12b): (1) 1 (1) (2) 0 11 )(x A G x x   . Since the operator 1 2G  is Lipschitz – continuous with Lipschitz coefficient equal to 1 , the substitution of the error 1 ( )q n into the right –hand side of the integral equation (12b) causes an error of not more than 1 ( )q n in the corresponding solution (1)x . The error of an iteration process in the calculation of (1)x equals 1 (2)1 11 n l q x q   . For any  1,2,..., ,l n we have   (2) (2) (2) (2) (2) (2) 1 0 01 (2) (2) (2)1 2 2 2 2 1 0 01 l l l l l x x x x x x